Finding inverse trig derivatives — Krista King Math | Online math help - Free Printable

The image you provided contains a table summarizing the derivatives of inverse trigonometric functions and their composite forms. Let's break it down systematically:

Inverse Trigonometric Functions and Their Derivatives

The table is divided into two main columns:
1. Basic Inverse Trigonometric Functions: These are the derivatives of the basic inverse trigonometric functions \( y = \sin^{-1}x, \cos^{-1}x, \tan^{-1}x, \cot^{-1}x, \sec^{-1}x, \csc^{-1}x \).
2. Composite Forms: These are the derivatives of inverse trigonometric functions applied to another function \( g(x) \), i.e., \( y = \sin^{-1}[g(x)], \cos^{-1}[g(x)], \tan^{-1}[g(x)], \cot^{-1}[g(x)], \sec^{-1}[g(x)], \csc^{-1}[g(x)] \).

Derivatives of Basic Inverse Trigonometric Functions

The first column lists the derivatives of the basic inverse trigonometric functions:

1. \( y = \sin^{-1}x \):
\[
y' = \frac{1}{\sqrt{1 - x^2}}
\]

2. \( y = \cos^{-1}x \):
\[
y' = -\frac{1}{\sqrt{1 - x^2}}
\]

3. \( y = \tan^{-1}x \):
\[
y' = \frac{1}{1 + x^2}
\]

4. \( y = \cot^{-1}x \):
\[
y' = -\frac{1}{1 + x^2}
\]

5. \( y = \sec^{-1}x \):
\[
y' = \frac{1}{|x|\sqrt{x^2 - 1}}
\]

6. \( y = \csc^{-1}x \):
\[
y' = -\frac{1}{|x|\sqrt{x^2 - 1}}
\]

Derivatives of Composite Inverse Trigonometric Functions

The second column extends these results to composite functions where the argument of the inverse trigonometric function is another function \( g(x) \). The chain rule is applied here.

1. \( y = \sin^{-1}[g(x)] \):
\[
y' = \frac{g'(x)}{\sqrt{1 - [g(x)]^2}}
\]

2. \( y = \cos^{-1}[g(x)] \):
\[
y' = -\frac{g'(x)}{\sqrt{1 - [g(x)]^2}}
\]

3. \( y = \tan^{-1}[g(x)] \):
\[
y' = \frac{g'(x)}{1 + [g(x)]^2}
\]

4. \( y = \cot^{-1}[g(x)] \):
\[
y' = -\frac{g'(x)}{1 + [g(x)]^2}
\]

5. \( y = \sec^{-1}[g(x)] \):
\[
y' = \frac{g'(x)}{|g(x)|\sqrt{[g(x)]^2 - 1}}
\]

6. \( y = \csc^{-1}[g(x)] \):
\[
y' = -\frac{g'(x)}{|g(x)|\sqrt{[g(x)]^2 - 1}}
\]

Explanation of the Chain Rule Application

For composite functions \( y = f[g(x)] \), the derivative is given by the chain rule:
\[
y' = f'[g(x)] \cdot g'(x)
\]
In the context of inverse trigonometric functions:
- For \( y = \sin^{-1}[g(x)] \), the derivative of \( \sin^{-1}u \) with respect to \( u \) is \( \frac{1}{\sqrt{1 - u^2}} \). Applying the chain rule:
\[
y' = \frac{1}{\sqrt{1 - [g(x)]^2}} \cdot g'(x) = \frac{g'(x)}{\sqrt{1 - [g(x)]^2}}
\]
- Similarly, for other functions, the derivative of the outer function is multiplied by the derivative of the inner function \( g(x) \).

Final Answer

The table summarizes the derivatives as follows:
\[
\boxed{
\begin{array}{lll}
y = \sin^{-1}x & y' = \frac{1}{\sqrt{1 - x^2}} & y = \sin^{-1}[g(x)] & y' = \frac{g'(x)}{\sqrt{1 - [g(x)]^2}} \\
y = \cos^{-1}x & y' = -\frac{1}{\sqrt{1 - x^2}} & y = \cos^{-1}[g(x)] & y' = -\frac{g'(x)}{\sqrt{1 - [g(x)]^2}} \\
y = \tan^{-1}x & y' = \frac{1}{1 + x^2} & y = \tan^{-1}[g(x)] & y' = \frac{g'(x)}{1 + [g(x)]^2} \\
y = \cot^{-1}x & y' = -\frac{1}{1 + x^2} & y = \cot^{-1}[g(x)] & y' = -\frac{g'(x)}{1 + [g(x)]^2} \\
y = \sec^{-1}x & y' = \frac{1}{|x|\sqrt{x^2 - 1}} & y = \sec^{-1}[g(x)] & y' = \frac{g'(x)}{|g(x)|\sqrt{[g(x)]^2 - 1}} \\
y = \csc^{-1}x & y' = -\frac{1}{|x|\sqrt{x^2 - 1}} & y = \csc^{-1}[g(x)] & y' = -\frac{g'(x)}{|g(x)|\sqrt{[g(x)]^2 - 1}}
\end{array}
}
\]